Design of Footings

 Lecture #07 

The Design of Spread Footings - De Desi sign gn Cr Crite iteri ria a - Des Design ign Pro Proced cedur uree - Example Proble Problem m for a Square Square Footing Footing - Exampl Examplee Problem Problem for a Rectang Rectangular ular Footing Footing - Example Problem for a Continuous Footing - Complex Cases

The Design Procedure. 1. Determine Determine the the structur structural al loads loads and member member sizes sizes at the founda foundation tion level level;; 2. Collect Collect all the geotec geotechnic hnical al data; data; set the proposed proposed footin footings gs on the geotechni geotechnical cal profile; profile; 3. Determine Determine the depth depth and locati location on of of all foundati foundation on element elements, s, 4. Dete Determ rmin inee the the beari bearing ng cap capac acit ity, y, 5. Determine Determine possibl possiblee total and and different differential ial settlemen settlements; ts; check check effects effects at 2B depth depths; s; 6. Select Select the the concret concretee strengt strength h (and (and possibly possibly the mix), mix), 7. Sele Select ct the the ste steel el grad grade, e, 8. Determ Determin inee the requ require ired d footi footing ng dimen dimensio sions, ns, 9. Determine Determine the foot footing ing thick thickness ness,, T (or D in some some textb textbook ooks), s), 10. Determine the size, size, number and spacing spacing of the reinforcing reinforcing bars, 11. Design the connection connection between the superstructure and the foundation, and 12. Check uplift and stability stability against sliding and overturning overturning of the structure-soil system. system. The first studies performed on foundation foundation structural failures failures were done by Professor Talbot at the University of Illinois in 1913. Advances in the next 50 years include Prof. F.E. Richart’s tests at the University of Michigan. His results were synthesized into the methodology used today by a committee sponsored sponsored by the ACI and ASCE and published in 1962. Spread footings is still the most popular foundation around the world because they are more economical than piles, adding adding weight to them does not not affect any other member, and and their performance has been excellent.

Selection of materials. Spread footings footings are usually usually designed to to use 3 ksi ksi 1.2). b) Wide beam shear for rectangular footings (that is, when L when  L /   /  B

 Analysis. The analysis of a square or rectangular footing may first be performed by assuming there is no steel in the member. The depth d  depth d from from the top of the footing to the tension axis is,

S  F y = 0 Qu = 2 d v c (b + d) + 2 d v c (c + d) + (c+d)(b+d) q o (shear on 4 faces) (bottom face) Set Qu = BL q o ∴

 d 2(4 v c + q o) + d   BL - cb) q o = 0 + d (2 (2 vc + q o)(b + c) - ( BL

column , wher For the special case of a square a square column, wheree c = b = w, w,  d 2 (v c + q o /4) + d  + d ((v c + q o /2)  /2)w w - ( B2 - w2) q o /4 = 0 column, with a = diameter, For the case of a round a round column,  d 2 (vc + q o /4)  /4) + d  + d ((vc + q o /2)  /2) a - ( BL - A col ) q o  /4 = 0

 Design Steps. Step 1. Compute the footing area via B x L, L, for a square footing BxB footing  BxB = SQRT(Q SQRT( Q / q  / q all ) for a rectangular footing  BxL = Q / q  / q all  where Q is the critical load combination (not Qu). Step 2.  Find the soil reaction under ultimate structural loads to check bearing capacity.  BL find the "ultimate" contact bearing, q o = Qu /  BL and check that q o ≤ qu Step 3. Compute the shear in the concrete v c .  . footing, check for diagonal Case (a) for a square a square footing, for diagonal tension (punching shear), v all  = 4∅ SQRT ( f'  f' c) where

∅ = 0.85 for shear.

For example, for f' for f' c = 3000 psi, v all  = 186 psi

footing, check for wide beam shear, shear, Case (b) for a rectangular a rectangular footing, v all 

≤

2∅ SQRT( f' c)

where ∅ = 0.85 for shear.

For example, for f' for f' c = 3000 psi, v all  = 93.1 psi Step 4.  Find the effective footing depth d.  d via this method eliminates (Note that use of  d via eliminates the need to use steel for shear, which is used only for flexure. Use the appropriate equation from the Analysis Section. Step 5. Compute the required area of steel A s (each way) for bending (flexure). Bending moment /unit width  M  = q L 2 / 2 (for a cantilever beam)  M u = qult L 2 / 2 =

 d – a /2) ∅ A s f  y ( d –

 p, so that the maximum allowed percentage of steel is not exceeded. Check  p Step 6. Compute bond bond length, column bearing, bearing, and the steel area required required for dowels. Use as a minimum an A an A s = 0.005 A 0.005 A col  (usually with 4 equal bars). Step 7.  Draft the above information into a complete drawing showing all the details.

 Example #1. Design a square reinforced concrete footing for the following conditions: - The column column has has a DL = 100 kips, kips, a LL = 120 kips, kips, and and is a 15” 15” x 15” with with 4 #8 bars; bars; - The footing footing is upon a soil with with q all  = 4 ksf with with a FS=2.5; FS=2.5; use f’ use f’ c = 3000 3000 psi psi and ƒ y = 50 ksi. Solution. Step 1. Find the footing dimensions (for service loads).

qall

=

Q  B

∴

2

B=

Q qall

=

220 4

= 7.42 ft

Use B Use B = 7.5 ft. ft.

Step 2. Check the ultimate parameters (that is, the actual soil pressure q o under Qu) The ultimate contact pressure,

Qu

∴

= 1.4DL + 1.7 LL = 1.4 (100 ) + 1.7 (120 ) = 140 + 204 = 344 kips qo

=

Qu  B

2

=

344 (7.5)

2

= 6.1 ksf < 10 ksf for qu

OK

 

Step 3. Compute the allowable shear stress in the concrete.

vall

= 4ϕ

f 'c

= 4(0.85)

3000 = 186 psi = 26.8 ksf

 

Step 4. Find the effective footing depth d  depth d (in (in this case punching shear governs),

 + qo  + d  v + qo  w −  B2 − w2 qo = 0 )   c 2   ( 4 4       2   6. 1 6 . 1 6 . 1 1 5 1 5       2 2 d  4)(26.8) +  + d  26.8 + 2  12 −  7.5 −  12      =0   4 4           

d 2  4vc

which yields two solutions: d  solutions:  d 1 = 1.20 ft and d  and d  2 = - 2.46 2.46 ft. ft. Choose d = 1.20 ft

15”

B = 7.5’

15” 15” + d

As a check, use the modified equation for d, which is,

4d 2 + 2 ( b + c ) d − B 2

qo vo

=0

 d = 1.28 ft which yields two solutions: d  solutions:  d 1 = 1.28 ft and d  and d  2 = - 2.52 2.52 ft. ft. Choos Choosee d = a difference of only only 7%. Therefore Therefore use the largest d = 1.28’ = 15.4 in. in. Use d = 16 in. It is not necessary to check for wide-beam shear. B = 7.5’

Step 5. Compute the required flexural steel area As. The unit strip of the cantilever arm L’ arm  L’ is,

'=

( B − w) L 2

=

[7.5'− 2

15 12

]

= 3 .1 3

ft

 

15”

1’

 L’

The cantilever moment M is,

=

qo L '2 M 2

where but

=

(6.1)(3.13) 2 (12) 2

= 359

−

a    M u = ϕ As f y  d −   with ϕ = 0.9  2  

a=

 As f y

0.85 f 'c b

which yields 16 As

=

As ( 50 )

0.85 ( 3) (12 in.)

in

kips

for flexure

= 1.63 As

− 0.81 As − 7.97 = 0 whence As = 0.50 in2 /

ft   

Total A Total A s = 7.5 ft (0.5 in 2 /ft) = 3.75 in 2  bd = 0.5 / (12) (16) = 0.0026 > 0.002 (minimum) Check ρ = A s /  bd = < 0.021 (maximum) In B In B = 90 in (7.5’)

 As = 3.72 in 2) @ 7.5” use 12 #5 bars ( As 7.5” cc  As = 4.20 in 2) @ 12” cc or 7 #7 bars ( As  As = 3.85 in 2) @ 18” cc or 5 #8 bars ( As

OK

Step 6. Check the development length for bond Ld (ACI 12.2, 12.6).  A b = the area of each individual bar, and d  If  A and d  b = the diameter of bar d  =

( 0.04 A f  ) L

b

 f  'c

y

=

[0.04(0.60)(50000)] 3000

= 21.9 > 12 in (mininmum)

also check with L with L d  = 0.0004 d 0.0004 d b f  y = 0.0004 (0.875 in.) (50,000 psi) = 17.5 inches. Step 7. Check the column bearing to determine need of dowels. Area of the column A column A g = w2 = (15 in.)2 = 225 in 2 Effective area of footing A footing  A 2 = (b +4d)2 = [15 in + 4(16 in) ]2 = 6,240 in 2 Checking the contact stress between column and footing,

 A2

c

=f ϕ

c

=f (0.7)(3000)(2) = 3750

'c f

 Ag

wheϕ re= 0.70

A2 a nd   Ag

p>s3i 000

≤2 psi

Need dow.els

The actual ultimate contact pressure at ultimate loads is,  f  c = Qu /  A  A g = 344 kips / 225 in 2 = 1.52 ksi < 3.0 ksi However, dowels However, dowels are always required , with at least A least A s

OK

≥ 0.005 A 0.005 A g

 A s required  = 0.005 (225 in 2) = 1.125 in 2 The diameter φ of the dowels are ± φ column bars ≤ 0.15” 0.15” (maximum (maximum diameter diameter difference) difference) Use 4 #7 bars or (4 x 0.60 in 2) = 2.40 in 2 = A s dowels, whereas column has #8’s Chec eck k di diamet meter di difference ence = 1. 1.00” – 0.87 .875” = 0.1 0.12 25 < 0.15 .15”

OK

Step 8. Check the embedment length L length L d  for the dowels.

L= d

d 

d 

0.2 f y d b  f  'c

or 0.0004

y

f bd or  8 inches

choose the largest of the three

0.2(50,000)(0.875) =L = 16 inc>h8es inches OK   3000 = L0.0004(50,000)(0.875) = 18 inc>h8es inches OK  

Note however that d  that d = = 16 in., need a 2” hook to reach 18”, 18”, but minimum minimum hook hook is 6”. Step 9. 9. Sketch the results.

 N =  N = 220 kips Qu = 344 kips 4 #7 dowels

19”

3” clearan clearance ce

B = 7.5’ 7 #7 #7 @ 12” 12” c-c each each way way

 Example #2. Design a rectangular footing footing to carry a moment induced by wind, with the following data, DL = 800 kN, LL= 800 kN, kN, M= 800 kN-m kN-m and q all  = 200 kN/m2 with FS = 1.5. Square colu column mnss with with c = 500 500 mm, mm, f'c = 21 MPa MPa and fy fy = 415 MPa. Step 1. Find the footing dimensions. This time, perform a trial and error selection of B x L. Set B Set Bx x L = B2 and check the increase in the soil pressure due to wind load.  / q all  = 1600 1600 kN / 200 kN/m kN/m2 = 8 m2  B2 = Q / q ∴  B

= 2.82 m

e = M  / Q  / Q = 800/1600 = 0.5

∴ L ≥ 6(0.5) = 3 m

from (6e/L)

 L = 3 m try If  L try a foot footin ing g 2.5 2.5 m x 4 m. m.

q avg = Q / A = 1600 1600 kN / 10 m2 = 160 kPa q max = Q /  BL  BL [ 1 + 6e /   L]  L] = (1600 /10)[1 + 6(0.5)/4] = 280 kPa Note that q max exceeds q avg by 33% , ∴ increase size to 2.75 m x 4.5 m. q avg = Q / A = 1600 / (2.75)(4.5) = 130 kPa q max = 130[1 130[1 + 6(0.5)] 6(0.5)] = 217 217 kPa kPa < 200 x 1.33 1.33 = 266 266 kPa kPa Iterate one again and settle with B with B = 3 m and L = 5 m. Step 2. Check the ultimate parameters (that is, the actual soil pressure q o under Qu).  Pu = 1.4DL + 1.7LL = 1.4(800) 1. 4(800) + 1.7(800) = 2480 kN  M u = 1.4DL + 1.7LL = 1.4(300) + 1.7(500) = 1270 kN-m  Pu = 1270 / 2480 e = M  /   /  P 2480 = 0.152 q max = ( Pu / A){1 A){1 + 6e/L] 6e/L] = (2480/1 (2480/15)[1 5)[1 + 6(0.152)/5 6(0.152)/5]] = 266 kPa kPa < qu = 300 kPa OK q min = ( Pu /A)[1  /A)[1 - 6e/L] = 2480/15[1 2480/15[1 - 6(0.152)/ 6(0.152)/5] 5] = 64.5 kPa q avg = 248 2480 0 / 15 = 165 165 kPa kPa < q all  = 200 kPa OK

Step 3. Compute the allowable shear stress in the concrete. The diagonal tension for f' for f' c = 21 MPa, v c = 1.29 MPa Step 4. Find the effective footing depth d  depth  d . Using the simplified equation, 4 d 2 + 2(b + c) d  c) d – – BL q o / v  / v c = 0 4 d 2 + 2(0.5m + 0.5m) d  0.5m) d -- (15)(165) (15)(165) / 1290 1290 = 0 which which yields yields d ~ 0.50 m Now find the depth d  depth  d for for wide beam, from x = 0 to x = 2.25 – d  dv = q dx V = V = ? qdx = ((26 266 6 - 40.2 40.2x) x)dx dx = {266x {266x – (40.2)(2 (40.2)(2)/2} )/2}  d - 20.1( = 598 598 - 266 266 d 20.1(2.2 2.25 5 - d )2 Vc = v c /2 d =  d = 1290 / 2 d   d = 0.60 m ∴ use the highest, d = 0.60 m. ∴  d =

Step 5. Compute the required longitudinal flexural steel area A  . area A s .

a

=

 As f y

0 . 8 5 f 'c b

=

2.25

M = u

∫

Mu

0 . 85( 2 1) (1 )

= 23.3 As

2.25

V d= x

0

∴

4 1 5 As

= 597

∫  [ 2 6 6

−x ( 4 0 . 2 ) ( 2 ) / 2 ]

d= x 597

kN −

m

0

k N − m = ϕ As fy ( d − 

2 A − 0.0515 s

A− 0 . 0 1 3 7 s

∴ use 1 7 # 2 5

mm rebars

=0

a

2

)

f r o m w h e n c e sA= 2 8 . 2

c 2m/

 m

 bd  = (0.00282)/(1)(0.6) = 0.0047 > 0.002 (minimum) Check ? =  As /  bd  < 0.021 (maximum) OK

area A s . Step 6. Compute the required transverse flexural steel area A  . Using a high average q = v( v (q avg.+ q max) = v(165 + 266) = 215 kPa Mu = wl2/2 = (q[(b-0.5)/2]2)/2= (q[(b-0.5)/2]2)/2= 215(215[(3-0.5)/2]2)/2 = 168 m-KN and again again As(d As(d - a/2) = Mu/  Mu/ ∅fy As2 - 0.0515As 0.0515As + 0.0003 0.000386 86 = 0 ∴ As = 7.61 cm2/m  bd = 7.61 x 10-4 m2  / (1)(0.6) = 0.00126 < 0.002 less than minimum check ? = A s /  bd = ∴ use minimum A  bd = 0.002(1)(0.6) = 0.0012 m2 /m = 12cm2 /m minimum A s = 0.002 bd =

Therefore, the longitudinal steel, 28.2 cm2 /m x 3m = 84.6 cm2 x 1 m2 /10,000cm2 = 8.46 x 10-3 m2 ⇒ 17 # 25 mm bars ∴ placed at 17.6 cm c-c. The transverse steel, 12 cm2 /m x 5m = 6.0 x 10-3 m2 ∴ placed at 38 cm c-c. Step 7. Draft a sketch of the results.

⇒ 13 # 25mm bars

 Example #3. Design continuous continuous wall footing for a warehouse building, building, given DL = 3 k/ft, LL = 1.2 k/ft, k/ft , the q all  = 2 ksf (with (with FS = 2), ?soil = 110 pcf, and f' and f' c = 3000 psi, f  psi, f  y = 60 ksi, ?concrete = 150 pcf.

Step 1. Assume a footing thickness D = 1 ft. By ACI 7.7.3.1, the minimum cover In soils is 3”. Also assume using #4 bars (∅ 1/2") ∴ d = 12" 12" - 3" - 0.5"/2 0.5"/2 = 8.7 8.75 5“

Continuous, Continuous, or strip, or wall footings, are reinforced transversely to the direction of the wall. The footing acts, in essence, as two small cantilever beams projecting projecting out from under the wall and perpendicular to it, in both directions. The reinforcement is placed at the bottom, where the stresses are in tension (flexure). Although the main reinforcement is transverse to the axis of  the wall, there is also a requirement for longitudinal reinforcement to control temperature tempera ture shrinkage and concrete creep.

Step 2. Find the soil pressure q o under ultimate loads.  B = Q / q  B = 3 feet Estimate Estimate the footing footing width width B / q all  = 4.2 k / 2 ksf = 2.1 ft, ft, assume assume B ACI 9.2 required strength U = 1.4D + 1.7L U = 1.4(3) 1.7(1.2) = 4.2 + 2.0 = 6.2 kips kips The soil pressure at ultimate loads, q o = U / (B)(1) (B)(1) = 6.2 kips kips /(3ft)(1ft) /(3ft)(1ft) = 2.1ksf < qu = 4 ksf 

Step 3. Check the shear strength of the concrete. The critical section for shear occurs at a distance d  distance d from from the face of the wall (ACI 15.5 and 11.11.1.2).

the ultimate shear Vu = (12" - 8.75") (1ft) (1ft) ft / 12 12 in x 2.1 k/ft k/ft2 = 0.57 kips /ft of wall Check the concrete strength, vu ≤ ∅ vc = 2∅ v( f'c)  f'c) bd  = (0.85)(2)v(3000)(12”)(8.75 (0.85)(2)v(3000)(12”)(8.75)) = 9.8 kips/ft of wall OK Since vu 6" ( d 

rechecking:  f'c) bd =  bd = 2(0.85)v(3000) (12in)(6.50in) Vu < ∅Vc = 2∅ v( f'c) (12in)(6.50in) = 7.3 kips/ft of wall OK Step 4. Compute the required transverse flexural steel area As. From ACI 15.4  M u = q o l  2 / 2 where l = l = 12” 12” = (2.1 ksf/2)(1ft ksf/2)(1ft2) = 1.05 k-ft /ft of wall

but a = A s f  y / 0.85 f' 0.85 f' c b = (60 ksi) As ksi) As /  / (0.85)(3 ksi)(12 in) = 1.96 A 1.96 A s (inches)  d - a/2) b = 12" of wall, and M  and M  n = A s f  y ( d a/2) = A s (60 k/in2)(6.5" - a /2) but M  but M u = ∴

∅ M  n = 0.9 M  0.9 M  n

1.05 kips-in / in = 0.9(60 k/in 2) A s (6.5 (6.5"" - 0.98 0.98 A s)

53 As2 - 351 As - 1.05 = 0 which which yield yieldss A s1 = 6.6 in2 per ft. of wall  A s2 = 0.003 in 2 per ft of wall ?1 = A s1 /  Bd   Bd  = 6.6 in 2 / (12 in)(6.5 in) = 0.085 ? 2 = A s2 /  Bd   Bd  = 0.003 in 2 / (12in)(6.5in) (12in)(6.5in) = 0.0004 0.0004 < 0.0018 The maximum steel percentage allowed ? max = 0.75 ? b where, ? b = 0.85( f' c /   f  y) ß [87,000/(87,000 + f   /  + f  y)] = 0.85(3/60) 0.85 (87,000/(87,000 + f  + f  y) = 0.021

∴? max

= 0.75 ? b = 0.75(0.021) = 0.0016 Now note that ?1 = 0.085 > ?max = 0.016 therefore use ? min = 0.0018. ∴

 bd = (0.0018)(12 A s = ? min bd = (0.0018)(12 in)(6.5 in) = 0.14 in2 per ft. of wall

∴

use 1 #4 every foot of wall  ( A  A s = 0.20 in 2)

Step 5. Check the reinforcement development length L length L d  (ACI 12.2).  L d  = 0.04 Ab fy / v( f' c) (but not less than 0.0004 d 0.0004 d b f  y)  L d  = 0.04 (0.20 in 2)(60,000 psi) psi) / v(3000psi) v (3000psi) = 8.8 inches  L d  = 0.004 d b fy = 0.004 (6.5 in)(60,000) Check  L in)(60,000) = 12 inches (12” controls controls design) design) The depth depth is thus thus 12" 12" - 3”(cover) = 9” < 12" 12" (thus (thus are are missing missing 3" on each side for L for L d ) ∴

must increase footing to B to B = 3.5 feet.

Minimum steel in longitudinal longitudinal direction to offset shrinkage shrinkage and temperature effects (as per ACI 7.12),  A s = (0.0018) b d = 0.0018(42 in)(6.5 in) = 0.49 in 2. (0.0018) b d = ∴

Provide 3 #4 bars at 12" ( A s = 0.60 in 2)

Step 6. Sketch the finished design.

 More Complex Cases - Large Billboar Billboard d Signs Signs or Highway Highway Signs

Cantilever sign are more economical than the overhead (bridge) type. They are built from hot-dipped galvanized steel pipes to a maximum truss span of 44 ft (13.4 m). The single-column steel post is bolted to a drilled shaft foundation. Design follows AASHT AASHTO’s O’s “Standard Specifications Specifications for Structural Supports for Highway Signs, Luminaires, and Traffic Signals”, published in 1985. This NJDOT research structure tended to move vertically and horizontally with the passing of trucks underneath. Anchor bolts loosened

In addition to low natural frequencies of about 1 Hz, these structures have extremely low damping ratios, typically 0.2 to 0.5 %. 1 percent. Cantilever Cantilever support structures are susceptible to large-amplitude large-amplitude vibration and fatigue cracking caused by wind loading. Report 412 of the National Research Program (NCHRP) “Fatigue “Fatigue Resistance Design of Cantilevered Signal, Sign, and Light Light Supports” Supports” studies three three phenomena: (1) buffeting by natural wind gusts, (2) buffeting by trucks, and (3) galloping (also known as “Den Hartog instability”. instability”. Galloping is an aeroelastic phenomenon phenomenon caused by wind generated aerodynamic forces. forces. Maximum response response is from truck gusts, gusts, as an equivalent equivalent static pressure of 37 psf ( 1,770 Pa) times a drag coefficient of 1.45 for vertical movement, and 1.7 for wind, in the horizontal direction. Full strength from truck-induced gusts occurred at 18 ft heights and decreased to zero at at 30 ft heights. (CE Sept. 2000).

 Homework: the billboard sign. Design a spread footing footing using FBC FBC and ASCE 7- 02. Ignore the effect of the water table. 32'

20'

P = 10 k  24" STEEL COLUMN 40 ' HIGH WITH 1" THICK WALLS

20' Z Y D B

X

STEP 1: Find Find the wind wind load as per ASCE 7-93, assuming an Exposure Exposure C, Cat. I.F = qz Gh Cf  AfTh AfThee force force wher whereqz eqz = 0.00 0.00256 256 kz (IV)2 (IV)2 ∴ qz = 34 psfan psfandKz dKz = 0.98I 0.98I = 1.05 1.05 V = 110 mph GH = 1.26CF = 1.2 The sign shape factor is M/N = 32/20 = 1.6 ∴ F = (34 psf) (1.26) (1.2) (32ft x 20ft) / 1000 = 32.4 kipsCalculate loads on footing footing.. STEP 2:Weigh 2:Weightt of steel column column = gs gs L A = 0.49 0.49 Mx Mx = 10 kips kips x 15’ 15’ = 150 k-ftM k-ftMy y = 32.4 k x 30’ = 972 k-ftMz k-ftMz = 32.4 k x 15’ 15’ = 486 k-ftTot k-ftTotal al (normal) (normal) load N = 10 10 k + 5 k = 15 kips kips STEP 3:Calculate 3:Calculate the footing’s footing’s bearing capacity capacity using Hansen’s Hansen’s formula.qult formula.qult = c Nc sc dc Ic +`q Nq sq dq Iq + 0.5 0.5 g B’ B’ Ng sg dg Igwher Igwherec ec (coh (cohes esio ion) n) = 0.1 0.150 50 ksfq ksfq = gD gDf = (em (embe bedm dmen entt pressure) pressure) = (0.130 (0.130 ksf)(3 ksf)(3 ft) = 0.39 ksfB ksfB = (footi (footing ng width width – initial initial assumpti assumptions) ons) = 5 ftL = (footing (footing length length – initial assumptions) = 15 ftNq (bearing cap. factor for embedment at f = 20°) = ep tan f tan2(45+f/2) = 6.40Nc (bearing cap. factor for cohesion at f = 20°) = (Nq – 1) cot f = 14.83Ng (bearing cap. factor for width at f = 20°) = 1.5 1.5 (Nq (Nq – 1) tan tan f = 2.95s 2.95sq q= (shape factor for embedment) = 1.0 + (B/L) sin f = 1.11sc = (shape factor for cohesion) = 1.0 + (Nq/Nc) (B/L) = 1.14sg = (shape (shape factor for width) = 1.0 – 0.4 (B/L) = .867dq = (depth factor for embedment) = 1 + 2 tan f (1 – sin f)2 (Df/B) = 1.19dc 1.19dc = (depth factor for cohesion) cohesion) = 1.0 + 0.4 (Df/B) = 1.24dg = (depth factor for width) = 1.0 ic ic = (inclination factor) factor) = 0.5 – √ (1 – H / (Af (Af Ca) ) where where ca = (0.6 (0.6 to 1.0) 1.0) ciq ciq = [ 1 – (0.5 (0.5 H) / (V +Af +Af Ca cot f)]d f)]d wher wheree 2 ≤ d ≤ 5ig = [1 – (0.7 (0.7 H) / (V +Af +Af Ca cot f)]a f)]a

Assume FS = 3Therefore, qult qult = (0.15 ksf) (14.83) (1.14) (1.24) (1.0) + (0.39)(6.40)(1.11)(1.19)(1.0) (0.39)(6.40)(1.11)(1.19)(1.0) + (0.5) (0.130 ksf) (5) (2.95) (.867) (1.0) = 7.27 and qall qall = qult qult / FS = 2.4 2.42 2 #3 Assume: Assume:B B = 10’SC = 1.0 + (0.431 x 0.2) = 1.09L = 50’DC 50’DC = 1.0 + (0.4)(3/10) (0.4)(3/10) = 1.12D = 3’Q = (130)x (130)x3 3 = 390B/L 390B/L = 0.2SQ 0.2SQ = 1.0 + 0.2 sin sin 20 = 1.07FS 1.07FS = 3.0SJ 3.0SJ = 1.0 – 0.4(0.2) 0.4(0.2) = 0.92DQ = 1 + (0.315)(3/10) = 1.09 QULT = (150) (14.83) (1.09) (1.12) + (390) (390) (6.4) (1.07) (1.09) + (0.5) (130) (10) (2.9) (0.92) (0.92) (1.0) = 7361 psf  Qa = qULT qULT / FS = 245 2453 3 psf psf = 2.5 2.5 ksf  ksf  QMAX,MIN QMAX,MIN = P/BL +- 6pey/b2l 6pey/b2l +- 6Pex/BL 6Pex/BL2 2 P = 15 kips + (3x10x50x0.150) / FTG WT. = 240 kips Ex = My/P = 972/240 = 4.05’ Ey = Mx/P = 150/240 150/240 = 0.625’ 0.625’ Qmax,m Qmax,min in = 240kips 240kips/500 /500 +- 6(240)(0. 6(240)(0.625) 625)/102 /102 x 50 +- 6(240)(4.0 6(240)(4.05)/1 5)/10x50 0x502 2 = 0.89 0.89 ksf < 2.5 = 0.067 OK 6x4.05/50 + 6x0.625/10 = 0.86 < 1.0 ok (e in middle 1/3) #3 CHECK OT (LONG DIRECTION) MOT = 32.4 KIPS (30+3) = 1069 KIP FT MR = 5 KIPS KIPS x 25’ 25’ + 225 KIPS x 25’ = 5750 5750 KIP FT F.S. = 5750 5750 / 1069 = 5.4 >> 1.5 OK CHECK SLIDING RS = S V tan f + CB = 240 tan f + 150(10) = 1587 KIPS >> 32.4 KIPS

#3 LOAD COMBINATION = 0.75 (1.4D + 1.7L + 1.7W) = 1.05D + 1.275W FACTORED LOADS: PU = 1.05 x 240 KIPS = 252 KIPS MUX = 1.05 x 150 KIP FT = 158 KIP FT MUY = 1.275 x 972 KIP FT = 1239 KIP FT THEREFORE QMAX,MIN = 336/10 x 50 ± 6 x 336 x 0.625 / 102 x 50 ± 6 x 336 x 4.9 / 10 x 502 = 0.989 KSF, 0.019 KSF CHECK BEAM SHEAR D = 36” – 4” = 32”

VU = 0.019x(50 0.019x(50 – 21.33) 21.33) + 0.15(50-21.33 0.15(50-21.33)(0.9 )(0.989 89 – 0.019) 0.019) = 14.4 K/FT K/FT FVC = 0.85 x 2 x √(3000) x 12 x 32 / 1000 = 36 KIPS > 14.4 KIPS OK PUNCHING SHEAR WILL NOT GOVERN BY OBSERVATION #3 DESIGN FOR FLEXURE IN LONG DIRECTION F’C = 3000 PSI FY = 60000 PSI MU = 0.019 0.019 x 26’ x 26’/2 + 0.5 x 26 x (0.989 (0.989 – 0.019) 0.019) x 26 / 3 = 116 116 KIP FT A = As x FY / 0.85 .85 F’C B AS = MU x 12 / F FY (D – A/2) A/2) A = 1.64”, AS = 0.83 IN2 R = AS /BD = 0.83 / 12 x 32 = 0.0022 > 0.0018 OK USE #7 #7 @ 8” 8” O/C, AS = 0.90 IN2 OK OK THEREFORE MU MU = 0.9 x 60 (132 (132 – 1.64/2)/12 = 126 KIP FT > 116 KIP FT OK USE FOOTING 10 x 50 x 3 w/ #7 @ 8”o.c. T&B E.W. NOTE: IN LIEU OF SUCH A LARGE (EXPENSIVE) FOOTING, A DRILLED SHAFT SH AFT MIGHT BE RECOMMENDED IN PRACTICE

 Footing Design Using Software

Shallow foundations are the most commonly used type of foundations, primarily because they are simple and very economical to build. Similarly, their design is simple, albeit tedious. Fortunately, the average price of software for the design and analysis of  shallow foundations is about $ 400. An example of an even simpler software is the the use of Microsoft’s® Excel spreadsheet. Given the loading conditions, the soil properties, and the footing’s material properties, a spreadsheet can give the dimensions of the footing and the maximum moment and shear acting upon it.

 Input Required.  a)  a) Phys Physic ical al Lay Layou outt : (1) the desired ratio of length to width, and (2) the estimated thickness of  the footing.  b) Material Properties: (1) Unit Weight of Concrete, and (2) Allowable Soil Bearing Capacity.  c) Loading Conditions: (1) Dead Load, (2) Live Load, (3) Moments about the x-axis and yaxis, and (4) Dead Load Imposed on Footing.

Computer Programs. SFOOTING is a program for the analysis and design of reinforced concrete spread footing foundations foundations subject to loading. The program allows the user to specify safety factors for geotechnical design and provides default minimum values. SFOOTING produces a design instantaneously and can thus be used to quickly study the effects of changes in footing footing depth, safety factors, loads and material strength properties.

 References. Braja M. Das, “Principles of Foundation Foundation Engineering”, Engineering”, 4th Edition. Joseph E. Bowles, “Foundat “ Foundation ion Analysis and Design” Design”,, 4th Edition. Arthur H. Nilson, Nilson, “Design “ Design of Concrete Structures, 12th Edition. Edward G. Nawy, “Reinfor “ Reinforced ced Concrete”, Leonard Spiegel & George F. Limbrunner, “ Reinforced Reinforced Concrete Concrete Design”, 4th Edition.